Can someone please verify my proof or offer suggestions for improvement? I am aware that there are similar questions posted elsewhere, but I need help with my proof in particular.
Some preliminaries:
Let $P$ be a poset. An antichain $C$ is a subset of $P$ such that for all $x, y \in C$ with $x \neq y$, $x$ is incomparable with $y$. An order ideal $I$ is a subset of $P$ such that $x \in I$ and $y \leq x$ implies $y \in I$.
Let $P$ be a finite poset. Show that the number of order ideals equals the number of antichains.
Let $I$ be an order ideal, and let $M$ be the set of maximal elements of $I$. Then, $M$ is an antichain in $P$ (Let $x, y \in M$ with $x \neq y$. If $x \leq y$, then $x$ is not a maximal element of $I$, a contradiction.) Let $I_1$ and $I_2$ be two order ideals in $P$ with the same set of maximal elements $M$. We show that it must be the case that $I_1 = I_2$. Suppose, for the sake of contradiction, that there exists an $x \in I_2$ such that $x \notin I_1$. Then, it must either be the case that $x \leq m$ for some maximal element $m$, or $x$ is incomparable with every maximal element. The first case implies that $x \in I_1$, and the second implies that $x$ is a maximal element of $I_2$. In either case, we arrive at a contradiction. Therefore, it must be the case that $I_1 = I_2$.
Now, let $A$ be an antichain in $P$. Set $I_A = \{x \in P: x \leq y$ for some $y \in A\}$. Then, $I_A$ is clearly an order ideal in $P$. Also, if $A \neq A'$, $I_A \neq I_{A'}$. Since there is a one-to-one correspondence between the order ideals and the antichains in $P$, and $P$ is finite, it must be the case that the number of order ideals and antichains in $P$ are equal.
Yes. The proof looks just fine.